The Anderson's bridge is an alternating current bridge, which is used to find the self-inductance of the circuit. Actually, this bridge is the modified form of Maxwell's inductance capacitance bridge for the measurement of self-inductance in terms of the standard capacitor of low Q coils (Q < 1). The bridge uses a fixed capacitor instead of a variable capacitor as seen in Maxwell's bridge and hence gives more accurate results.

## Construction of Anderson’s Bridge :

The bridge consists of four arms AB, AD, BC, and CD. The arm AB of the bridge consists of unknown inductance to be measured in series with a non-inductive resistance R_{1} and r_{1} be the resistance of the inductor. The below shows the circuit diagram of Anderson's bridge under balance condition.

The arms AD, BC, and CD consist of standard non-inductive resistances R_{2}, R_{3,} and R_{4} of known value. It consists of extra junction E to which a variable resistor r is connected between E and D, a fixed capacitor between E and C, and a null indicator or detector between E and B.

## Operation and Theory of Anderson’s Bridge :

#### Let,- C = Standard capacitor of fixed known value
- R
_{2}, R_{3}, R_{4} = Standard non-inductive resistances of known values - L
_{1} = Self-inductance to be measured - r
_{1} = Resistance of the unknown inductor.

Under balanced condition, we have,
From equation 3, we have,
From equation 4, we have,
From equation 5, we have,
From equations 6 and 7, we have,
Equating real and imaginary parts on both sides, we get,
Therefore, the unknown value of self-inductance is,
The unknown value of resistance of the self inductor is,

_{2}, R_{3}, R_{4}= Standard non-inductive resistances of known values_{1}= Self-inductance to be measured_{1}= Resistance of the unknown inductor.## Phasor Diagram of Anderson’s Bridge :

Taking current I_{1} as the reference phasor. The current I_{1} is passing through arm AB i.e., through non-inductive resistance R_{1} and inductor L_{1}. Thus the total drop V_{1} in arm AB is I_{1} (R_{1} + r_{1}) (which is in phase with I_{1}) and I_{1} ωL_{1}.

When the bridge is balanced the same current will also pass through R_{3} (in arm BC) i.e., I_{1} = I_{3}. Thus the drop I_{3} R_{3} (i.e., V_{3}) lies along the phasor I_{1}. Also, we know that the potential between junctions BC will be equal to the potential between junctions EC. Hence V_{3} = I_{c}/ωC and the drop I_{c} r in the resistor r will lie along with I_{c}.

When the potential across the detector is zero (balanced condition) V_{4} will be equal to V_{EC}. Thus V_{4} is nothing but the sum of drop, in resistor r, and V_{EC} or V_{3}. Adding the drop I_{c} r to the V_{3} we get V_{4} (i.e., drop I_{4} R_{4} in arm CD), also the current I_{4} lies along with V_{4}.

Also, when the bridge is at balance condition, the current I_{2} is the sum of the currents I_{c} and I_{4}, also the drop V_{2} (i.e., I_{2} R_{2}) in arm AD due to I_{2} lies along with it. Therefore, the resultant of the phasors V_{1} and V_{3} or V_{2} and V_{4} gives supply voltage V.